Biology Reference
In-Depth Information
Now, because
dA
1
b
dA
dt
dt
¼b
A
ð
t
Þ
, we have A
ð
t
Þ¼
and:
Z
1
Z
1
1
0
¼
1
b
1
b
1
b
A
0
b
:
A
ð
t
Þ
dt
¼
dA
ð
t
Þ¼
A
ð
t
Þ
A
ð
0
Þ¼
0
0
Thus, the average duration of the disease is:
0
1
0
1
Z
1
Z
1
Z
1
1
A
0
1
A
0
1
A
0
1
A
0
A
0
b
¼
1
b
:
d
@
A
¼
@
A
¼
¼
tdA
ð
t
Þ¼
A
ð
t
Þ
dt
A
ð
t
Þ
dt
0
0
0
3. The Long-Term Evolution of the Disease
As time passes and more susceptibles become infected and more
infectives recover, what is the long-term behavior of the disease?
Notice that because of the condition S(t)
þ
I(t)
¼
N, Eq. (2-2) of the model
could be written as:
¼ a
dI
dt
¼ a
b
a
b
a
IS
IN
I
:
b
Next, if N
a
>
0, we could further rewrite the right-hand side as
0
1
1
I
b
a
b
b
a
I
I
K
@
A
a
IN
I
¼ a
IN
a
I
¼ a
N
I
¼
r 1
;
b
a
N
¼ a
b
b
a
where K
¼
N
a
;
and r
¼ a
N
N
b >
0
:
Thus,
Eq. (2-2) takes the form:
1
I
dI
dt
¼
I
K
r 1
:
(2-3)
b
Does this equation look familiar? If K
0, the rate of change
for the group of infectives is given by a logistic equation! This also
means that we already know the long-term behavior for I(t), because in
Chapter 1 we studied the logistic equation in detail. Finally, knowing that
S(t)
¼
N
a
>
I(t) allows us to derive the long-term behavior of S(t) from
that of I(t). We present the results in the next two exercises.
¼
N
E
XERCISE
2-1
b
b
Show that for N
a
<
0
ð
or N
a
¼
0
Þ
, the number of infectives I(t)is
declining and, thus, there is no epidemic.
1. For Eq. (2-3), be sure to properly distinguish between the number 1 and the
symbol I that denotes the number of infectives.
